Pure 8.7 - Separable first-order differential equations | Edexcel A-Level Mathematics

A differential equation describes a function through its rate of change. You will separate variables, integrate to find a family of solutions, use a condition to choose one member, and check its branch, domain and sketch.

What separable means

If a curve has gradient zero at $(0,3)$ , must it stay horizontal? For $\mathrm{d}y/\mathrm{d}x=xy$ , the gradient is zero whenever $x=0$ , whatever the value of $y$ . Yet it can change immediately afterwards. The equation tells us the gradient at every point; one zero gradient is only one piece of the story.

You already know

Differentiating a function gives its gradient. Integration reverses differentiation and introduces an arbitrary constant. Use $\int1/y\,\mathrm{d}y=\ln|y|+C$ on an interval where $y\ne0$ ; this logarithmic integral is not in the formula booklet.

Before integrating, place each variable beside its own differential. Separation of variables and the use of an initial condition are not supplied in the formula booklet: learn the method.

First-order separable differential equation

A first-order differential equation contains a first derivative and no higher derivatives. It is separable when it can be rearranged, on an appropriate domain, as

F(y)\,\mathrm{d}y=G(x)\,\mathrm{d}x.

For example, $y'=3x^2y$ separates as $\mathrm{d}y/y=3x^2\,\mathrm{d}x$ for $y\ne0$ . A sum such as $y'=x+y$ does not separate by merely dividing each term by a different variable.

The separated line is shorthand for a valid chain-rule integration. We do not treat a derivative as an arbitrary fraction: after integrating $F(y)$ , the chain rule returns $F(y)y'$ , which explains the rearrangement.

Always inspect a factor before dividing by it. In $y'=3x^2y$ , the constant function $y=0$ satisfies the original equation, but division by $y$ would remove it. Record it and then solve the non-zero case.

Key idea

Separate products of single-variable factors, and check any constant solutions excluded by division.

Here is a local algebra check before the full method.

Finding a general solution

Once separated, integrate both sides. Two arbitrary additive constants can be combined into one, so a first-order general solution normally carries one free constant.

General solution

A general solution describes a family of functions using an arbitrary constant, on intervals where the original equation is defined. For example, different values of $A$ in $y=Ae^{x^3}$ give different solution curves. One fixed choice of $A$ is a member of the family rather than the general solution.

An additive constant inside a logarithmic equation becomes a multiplicative constant after exponentiating.

Worked example 1: an exponential family

Question. Find the general solution of $\mathrm{d}y/\mathrm{d}x=3x^2y$ .

Plan. Check $y=0$ , then divide by $y$ , integrate and absorb the sign into a new constant.

\begin{aligned} \frac1y\,\mathrm{d}y&=3x^2\,\mathrm{d}x && y\ne0\\ \ln|y|&=x^3+C && \text{integrate both sides}\\ |y|&=e^C e^{x^3} && \text{exponentiate}\\ y&=Ae^{x^3} && A=\pm e^C \end{aligned}

Why change the constant's name? $e^C$ is positive, whereas a solution may be positive or negative; $A$ absorbs that sign.

Answer. $y=Ae^{x^3}$ , with real $A$ . The value $A=0$ restores the separately checked constant solution.

Check. Differentiation gives $y'=3x^2Ae^{x^3}=3x^2y$ , including $A=0$ .

Change only the coefficient $3$ in the equation to $6$ . The separation and constant handling remain the same.

Worked example 2: a changed growth coefficient

Question. Find the general solution of $\mathrm{d}y/\mathrm{d}x=6x^2y$ .

Plan. Separate non-zero solutions and include the constant solution after checking the original equation.

\begin{aligned} \int\frac1y\,\mathrm{d}y&=\int6x^2\,\mathrm{d}x && \text{separate}\\ \ln|y|&=2x^3+C && \text{power rule}\\ y&=Ae^{2x^3} && \text{exponentiate and absorb sign} \end{aligned}

Why $2x^3$ rather than $6x^3$ ? Integrating $x^2$ divides by its new power $3$ .

Answer. $y=Ae^{2x^3}$ , for real $A$ , including $A=0$ .

Check. Its derivative is $6x^2Ae^{2x^3}$ , exactly $6x^2y$ .

Not every separated integral produces an exponential. For $y'=4x/y$ , where $y\ne0$ , separation gives $y\,\mathrm{d}y=4x\,\mathrm{d}x$ , so $y^2/2=2x^2+C$ , or $y^2=4x^2+C$ after rescaling the constant. This relation needs a non-zero branch on a suitable interval.

Watch out

Wrong: $\ln|y|=x^3+C\Rightarrow y=e^{x^3}+C$ . Right: $y=Ae^{x^3}$ , with the constant multiplying the exponential. Exponentiating a sum gives a product.

Change the coefficient once more.

Finding a particular solution

A point on the curve selects the constant. Substitute the initial condition into the integrated relation, not just into the differential equation: one gradient equation does not determine an entire curve.

Particular solution

A particular solution is a solution with its arbitrary constant determined by a condition. It must satisfy both the differential equation and that condition on its chosen domain.

First find the family, then identify the member you need.

Worked example 3: a positive initial value

Question. Solve $\mathrm{d}y/\mathrm{d}x=2xy$ , given $y=3$ when $x=0$ .

Plan. Separate, integrate and use the given point to determine the exponential multiplier.

\begin{aligned} \frac1y\,\mathrm{d}y&=2x\,\mathrm{d}x && \text{separate}\\ \ln|y|&=x^2+C\Rightarrow y=Ae^{x^2} && \text{integrate and exponentiate}\\ 3&=Ae^0\Rightarrow A=3 && \text{use the condition} \end{aligned}

Why use the condition after integration? It identifies a member of the family rather than just a gradient at one point.

Answer. $y=3e^{x^2}$ .

Check. $y'=6xe^{x^2}=2xy$ , and $y(0)=3$ .

Change only the initial value from $3$ to $-3$ . This checks why the modulus and the signed constant matter.

Worked example 4: a negative initial value

Question. Solve $\mathrm{d}y/\mathrm{d}x=2xy$ , given $y=-3$ when $x=0$ .

Plan. Use the same family $y=Ae^{x^2}$ , allowing a negative multiplier.

\begin{aligned} \ln|y|&=x^2+C && \text{integrated equation}\\ y&=Ae^{x^2} && \text{sign included in }A\\ -3&=Ae^0\Rightarrow A=-3 && \text{initial condition} \end{aligned}

Why not replace $\ln|y|$ by $\ln y$ ? The required solution is negative, so $\ln y$ would be undefined over the reals.

Answer. $y=-3e^{x^2}$ .

Check. $y'=-6xe^{x^2}=2xy$ , and the initial value is $-3$ .

A condition away from zero works in the same way: $y'=xy$ , with $y(1)=2$ , gives $y=2e^{(x^2-1)/2}$ . Keep the initial coordinate in the exponent until the constant is fixed.

Watch out

Checking only the differential equation is insufficient. Every member of the family does that; the initial condition chooses the required one.

Change only the initial condition for the equation in Worked example 4.

Factor before separating

The equation $y'=xy+2x$ contains a sum, but both terms share $x$ : write $y'=x(y+2)$ . This common factor exposes the separated structure. In contrast, $x+y$ has no such common-factor rearrangement.

Worked example 5: exposing a common factor

Question. Solve $\mathrm{d}y/\mathrm{d}x=xy+2x$ , given $y(0)=1$ .

Plan. Factorise, check the constant solution, then separate the non-constant case.

\begin{aligned} y'&=x(y+2) && \text{factor out }x\\ \frac1{y+2}\,\mathrm{d}y&=x\,\mathrm{d}x && y\ne-2\\ \ln|y+2|&=\frac{x^2}{2}+C && \text{integrate}\\ y+2&=Ae^{x^2/2},\quad 3=A && \text{use }y(0)=1 \end{aligned}

Why inspect $y=-2$ ? It makes the divisor zero and is a solution of the original equation, although it fails this initial condition.

Answer. $y=3e^{x^2/2}-2$ .

Check. $y'=3xe^{x^2/2}$ and $xy+2x=x(3e^{x^2/2}-2)+2x=3xe^{x^2/2}$ ; also $y(0)=1$ .

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Pure Mathematics (Papers 1 and 2)

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Mechanics (Paper 3, Section B)

Pure 8.7 - Separable first-order differential equations

What separable means

Finding a general solution

Finding a particular solution

Factor before separating

Continue this lesson.

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